3.134 \(\int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=69 \[ \frac{a \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{d}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} d} \]

[Out]

-((a^(5/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d)) + (a*Sec[c + d*x]^2*(a + a*Sin[c
+ d*x])^(3/2))/d

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Rubi [A]  time = 0.109976, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2676, 2667, 63, 206} \[ \frac{a \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{d}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((a^(5/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d)) + (a*Sec[c + d*x]^2*(a + a*Sin[c
+ d*x])^(3/2))/d

Rule 2676

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-2*b*
(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(p + 1)), x] + Dist[(b^2*(2*m + p - 1))/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2
- b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && IntegersQ[2*m, 2*p]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=\frac{a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac{1}{2} a^2 \int \sec (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{2 d}\\ &=\frac{a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{d}\\ &=-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} d}+\frac{a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}\\ \end{align*}

Mathematica [A]  time = 0.368171, size = 75, normalized size = 1.09 \[ \frac{a^2 \left (-\frac{2 \sqrt{a (\sin (c+d x)+1)}}{\sin (c+d x)-1}-\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a (\sin (c+d x)+1)}}{\sqrt{2} \sqrt{a}}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(a^2*(-(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]) - (2*Sqrt[a*(1 + Sin[c + d*x])]
)/(-1 + Sin[c + d*x])))/(2*d)

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Maple [A]  time = 0.107, size = 66, normalized size = 1. \begin{align*} -{\frac{{a}^{3}}{d} \left ({\frac{1}{a\sin \left ( dx+c \right ) -a}\sqrt{a+a\sin \left ( dx+c \right ) }}+{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x)

[Out]

-a^3*((a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)+1/2*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a
^(1/2)))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75117, size = 263, normalized size = 3.81 \begin{align*} \frac{\sqrt{2}{\left (a^{2} \sin \left (d x + c\right ) - a^{2}\right )} \sqrt{a} \log \left (-\frac{a \sin \left (d x + c\right ) - 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, \sqrt{a \sin \left (d x + c\right ) + a} a^{2}}{4 \,{\left (d \sin \left (d x + c\right ) - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/4*(sqrt(2)*(a^2*sin(d*x + c) - a^2)*sqrt(a)*log(-(a*sin(d*x + c) - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a
) + 3*a)/(sin(d*x + c) - 1)) - 4*sqrt(a*sin(d*x + c) + a)*a^2)/(d*sin(d*x + c) - d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out